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Binomial Theorem

Section: Mathematics

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Question : 32 of 33

Marks: +1, -0

The co-efficient of

x^5

(3+x+x^2)^6

[AP EAPCET 18 May 2024 Shift 1]

18
540
1620
2178

Solution:

We have,

(3+x+x^2)^6

General term of the expansion

=\;\frac{6!}{r!s!t!} \times 3^r \times x^5 \times (x^2)^t

Where,

\;r+s+t=6

\;=\;\frac{6!}{r!s!t!} \times 3^r \times x^{5+2t}

For the coefficient of

x^5

,

s+2t=5

But

r+s+t=6

\;r+5-t=6

\;r-a=1+t \; \text{ and } \; s=5-2t

Now,

t=0

, then

r=1, s=5

t=1

, then

r=2, s=3

t=2

then

r=3, s=1

\therefore

There are only 3 terms that contain

x^5

\therefore

Coefficient of

x^5

\;=\;\frac{6!}{0!1!5!} \times 3^1 + \;\frac{6!}{1!2!3!} \times 3^2 + \;\frac{6!}{2!3!!!} \times 3^3

\;=18+540+1620=2178

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